Math help please 10 point top answer

5*x*3*y=180
X^2*y^2=40
8x-8y=32
12x/4y=1

What's x and y but has to be the same and work for all of them... Also if it says for example z=2, 6z would be 62 not 6*2

Multiply both sides of the last one by 4y.
12x = 4y
Now multiply both sides by 2.
24x = 8y

Now you can substitute (24x) in for (8y) in the third one because we just proved they're equal.
8x - 8y = 32
8x - (24x) = 32
-16x = 32
x = -2

Now plug that back into the equation to solve for y.
8x - 8y = 32
8(-2) - 8y = 32
-16 - 8y = 32
-8y = 48
y = -6

So our (x,y) is (-2,-6). You can check by plugging those values into all 4 equations and verifying that they're true.

(The second one looks wrong. Could it be x^2 y^2 = 40? That would work.)

(12x) / (4y) = 1 => 3x = y, so

8x - 8y = 32 => 8x - 8(3x) = 32 => -16x = 32 =>/

/=> x = -2 => 3(-2) = y => y = -6

(x, y) = (-2, -6)

Surely, you meant x^2 y^2 = 40; otherwise, the values for x and y would not work for all of the equations.

statement z = 2 gives 6z = 62 not 12 doesn't conform to answers x = --2 and y = --6
as when y = --6 what is 4y? 4y = 4--6 = --2 not 12x which is 12--2 = 10
clarify the statement 6z = 62 when z = 2.